3.130 \(\int \frac {\sin ^{-1}(a+b x)}{x^5} \, dx\)

Optimal. Leaf size=186 \[ -\frac {a \left (2 a^2+3\right ) b^4 \tanh ^{-1}\left (\frac {1-a (a+b x)}{\sqrt {1-a^2} \sqrt {1-(a+b x)^2}}\right )}{8 \left (1-a^2\right )^{7/2}}-\frac {\left (11 a^2+4\right ) b^3 \sqrt {1-(a+b x)^2}}{24 \left (1-a^2\right )^3 x}-\frac {5 a b^2 \sqrt {1-(a+b x)^2}}{24 \left (1-a^2\right )^2 x^2}-\frac {b \sqrt {1-(a+b x)^2}}{12 \left (1-a^2\right ) x^3}-\frac {\sin ^{-1}(a+b x)}{4 x^4} \]

[Out]

-1/4*arcsin(b*x+a)/x^4-1/8*a*(2*a^2+3)*b^4*arctanh((1-a*(b*x+a))/(-a^2+1)^(1/2)/(1-(b*x+a)^2)^(1/2))/(-a^2+1)^
(7/2)-1/12*b*(1-(b*x+a)^2)^(1/2)/(-a^2+1)/x^3-5/24*a*b^2*(1-(b*x+a)^2)^(1/2)/(-a^2+1)^2/x^2-1/24*(11*a^2+4)*b^
3*(1-(b*x+a)^2)^(1/2)/(-a^2+1)^3/x

________________________________________________________________________________________

Rubi [A]  time = 0.27, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {4805, 4743, 745, 835, 807, 725, 206} \[ -\frac {5 a b^2 \sqrt {1-(a+b x)^2}}{24 \left (1-a^2\right )^2 x^2}-\frac {\left (11 a^2+4\right ) b^3 \sqrt {1-(a+b x)^2}}{24 \left (1-a^2\right )^3 x}-\frac {a \left (2 a^2+3\right ) b^4 \tanh ^{-1}\left (\frac {1-a (a+b x)}{\sqrt {1-a^2} \sqrt {1-(a+b x)^2}}\right )}{8 \left (1-a^2\right )^{7/2}}-\frac {b \sqrt {1-(a+b x)^2}}{12 \left (1-a^2\right ) x^3}-\frac {\sin ^{-1}(a+b x)}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[a + b*x]/x^5,x]

[Out]

-(b*Sqrt[1 - (a + b*x)^2])/(12*(1 - a^2)*x^3) - (5*a*b^2*Sqrt[1 - (a + b*x)^2])/(24*(1 - a^2)^2*x^2) - ((4 + 1
1*a^2)*b^3*Sqrt[1 - (a + b*x)^2])/(24*(1 - a^2)^3*x) - ArcSin[a + b*x]/(4*x^4) - (a*(3 + 2*a^2)*b^4*ArcTanh[(1
- a*(a + b*x))/(Sqrt[1 - a^2]*Sqrt[1 - (a + b*x)^2])])/(8*(1 - a^2)^(7/2))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 745

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1)*(a + c*x^2)^(p
+ 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[c/((m + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^(m + 1)*Simp[d*(m + 1)
- e*(m + 2*p + 3)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[
m, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]) || (SumSimplerQ[m, 1] && IntegerQ[p]) || ILtQ
[Simplify[m + 2*p + 3], 0])

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 835

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((e*f - d*g)
*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rule 4743

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*ArcSin[c*x])^n)/(e*(m + 1)), x] - Dist[(b*c*n)/(e*(m + 1)), Int[((d + e*x)^(m + 1)*(a + b*ArcSin[c*x])^(
n - 1))/Sqrt[1 - c^2*x^2], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 4805

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

Rubi steps

\begin {align*} \int \frac {\sin ^{-1}(a+b x)}{x^5} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\sin ^{-1}(x)}{\left (-\frac {a}{b}+\frac {x}{b}\right )^5} \, dx,x,a+b x\right )}{b}\\ &=-\frac {\sin ^{-1}(a+b x)}{4 x^4}+\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{\left (-\frac {a}{b}+\frac {x}{b}\right )^4 \sqrt {1-x^2}} \, dx,x,a+b x\right )\\ &=-\frac {b \sqrt {1-(a+b x)^2}}{12 \left (1-a^2\right ) x^3}-\frac {\sin ^{-1}(a+b x)}{4 x^4}+\frac {b^2 \operatorname {Subst}\left (\int \frac {\frac {3 a}{b}+\frac {2 x}{b}}{\left (-\frac {a}{b}+\frac {x}{b}\right )^3 \sqrt {1-x^2}} \, dx,x,a+b x\right )}{12 \left (1-a^2\right )}\\ &=-\frac {b \sqrt {1-(a+b x)^2}}{12 \left (1-a^2\right ) x^3}-\frac {5 a b^2 \sqrt {1-(a+b x)^2}}{24 \left (1-a^2\right )^2 x^2}-\frac {\sin ^{-1}(a+b x)}{4 x^4}-\frac {b^4 \operatorname {Subst}\left (\int \frac {-\frac {2 \left (2+3 a^2\right )}{b^2}-\frac {5 a x}{b^2}}{\left (-\frac {a}{b}+\frac {x}{b}\right )^2 \sqrt {1-x^2}} \, dx,x,a+b x\right )}{24 \left (1-a^2\right )^2}\\ &=-\frac {b \sqrt {1-(a+b x)^2}}{12 \left (1-a^2\right ) x^3}-\frac {5 a b^2 \sqrt {1-(a+b x)^2}}{24 \left (1-a^2\right )^2 x^2}-\frac {\left (4+11 a^2\right ) b^3 \sqrt {1-(a+b x)^2}}{24 \left (1-a^2\right )^3 x}-\frac {\sin ^{-1}(a+b x)}{4 x^4}+\frac {\left (a \left (3+2 a^2\right ) b^3\right ) \operatorname {Subst}\left (\int \frac {1}{\left (-\frac {a}{b}+\frac {x}{b}\right ) \sqrt {1-x^2}} \, dx,x,a+b x\right )}{8 \left (1-a^2\right )^3}\\ &=-\frac {b \sqrt {1-(a+b x)^2}}{12 \left (1-a^2\right ) x^3}-\frac {5 a b^2 \sqrt {1-(a+b x)^2}}{24 \left (1-a^2\right )^2 x^2}-\frac {\left (4+11 a^2\right ) b^3 \sqrt {1-(a+b x)^2}}{24 \left (1-a^2\right )^3 x}-\frac {\sin ^{-1}(a+b x)}{4 x^4}-\frac {\left (a \left (3+2 a^2\right ) b^3\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {1}{b^2}-\frac {a^2}{b^2}-x^2} \, dx,x,\frac {\frac {1}{b}-\frac {a (a+b x)}{b}}{\sqrt {1-(a+b x)^2}}\right )}{8 \left (1-a^2\right )^3}\\ &=-\frac {b \sqrt {1-(a+b x)^2}}{12 \left (1-a^2\right ) x^3}-\frac {5 a b^2 \sqrt {1-(a+b x)^2}}{24 \left (1-a^2\right )^2 x^2}-\frac {\left (4+11 a^2\right ) b^3 \sqrt {1-(a+b x)^2}}{24 \left (1-a^2\right )^3 x}-\frac {\sin ^{-1}(a+b x)}{4 x^4}-\frac {a \left (3+2 a^2\right ) b^4 \tanh ^{-1}\left (\frac {1-a (a+b x)}{\sqrt {1-a^2} \sqrt {1-(a+b x)^2}}\right )}{8 \left (1-a^2\right )^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.26, size = 194, normalized size = 1.04 \[ \frac {1}{8} \left (\frac {a \left (2 a^2+3\right ) b^4 \log (x)}{\left (1-a^2\right )^{7/2}}-\frac {a \left (2 a^2+3\right ) b^4 \log \left (\sqrt {1-a^2} \sqrt {-a^2-2 a b x-b^2 x^2+1}-a^2-a b x+1\right )}{\left (1-a^2\right )^{7/2}}+\frac {b \sqrt {-a^2-2 a b x-b^2 x^2+1} \left (2 a^4-5 a^3 b x+a^2 \left (11 b^2 x^2-4\right )+5 a b x+4 b^2 x^2+2\right )}{3 \left (a^2-1\right )^3 x^3}-\frac {2 \sin ^{-1}(a+b x)}{x^4}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[a + b*x]/x^5,x]

[Out]

((b*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]*(2 + 2*a^4 + 5*a*b*x - 5*a^3*b*x + 4*b^2*x^2 + a^2*(-4 + 11*b^2*x^2)))/(
3*(-1 + a^2)^3*x^3) - (2*ArcSin[a + b*x])/x^4 + (a*(3 + 2*a^2)*b^4*Log[x])/(1 - a^2)^(7/2) - (a*(3 + 2*a^2)*b^
4*Log[1 - a^2 - a*b*x + Sqrt[1 - a^2]*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]])/(1 - a^2)^(7/2))/8

________________________________________________________________________________________

fricas [A]  time = 0.59, size = 484, normalized size = 2.60 \[ \left [-\frac {3 \, {\left (2 \, a^{3} + 3 \, a\right )} \sqrt {-a^{2} + 1} b^{4} x^{4} \log \left (\frac {{\left (2 \, a^{2} - 1\right )} b^{2} x^{2} + 2 \, a^{4} + 4 \, {\left (a^{3} - a\right )} b x - 2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a b x + a^{2} - 1\right )} \sqrt {-a^{2} + 1} - 4 \, a^{2} + 2}{x^{2}}\right ) + 12 \, {\left (a^{8} - 4 \, a^{6} + 6 \, a^{4} - 4 \, a^{2} + 1\right )} \arcsin \left (b x + a\right ) - 2 \, {\left ({\left (11 \, a^{4} - 7 \, a^{2} - 4\right )} b^{3} x^{3} - 5 \, {\left (a^{5} - 2 \, a^{3} + a\right )} b^{2} x^{2} + 2 \, {\left (a^{6} - 3 \, a^{4} + 3 \, a^{2} - 1\right )} b x\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{48 \, {\left (a^{8} - 4 \, a^{6} + 6 \, a^{4} - 4 \, a^{2} + 1\right )} x^{4}}, -\frac {3 \, {\left (2 \, a^{3} + 3 \, a\right )} \sqrt {a^{2} - 1} b^{4} x^{4} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a b x + a^{2} - 1\right )} \sqrt {a^{2} - 1}}{{\left (a^{2} - 1\right )} b^{2} x^{2} + a^{4} + 2 \, {\left (a^{3} - a\right )} b x - 2 \, a^{2} + 1}\right ) + 6 \, {\left (a^{8} - 4 \, a^{6} + 6 \, a^{4} - 4 \, a^{2} + 1\right )} \arcsin \left (b x + a\right ) - {\left ({\left (11 \, a^{4} - 7 \, a^{2} - 4\right )} b^{3} x^{3} - 5 \, {\left (a^{5} - 2 \, a^{3} + a\right )} b^{2} x^{2} + 2 \, {\left (a^{6} - 3 \, a^{4} + 3 \, a^{2} - 1\right )} b x\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{24 \, {\left (a^{8} - 4 \, a^{6} + 6 \, a^{4} - 4 \, a^{2} + 1\right )} x^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(b*x+a)/x^5,x, algorithm="fricas")

[Out]

[-1/48*(3*(2*a^3 + 3*a)*sqrt(-a^2 + 1)*b^4*x^4*log(((2*a^2 - 1)*b^2*x^2 + 2*a^4 + 4*(a^3 - a)*b*x - 2*sqrt(-b^
2*x^2 - 2*a*b*x - a^2 + 1)*(a*b*x + a^2 - 1)*sqrt(-a^2 + 1) - 4*a^2 + 2)/x^2) + 12*(a^8 - 4*a^6 + 6*a^4 - 4*a^
2 + 1)*arcsin(b*x + a) - 2*((11*a^4 - 7*a^2 - 4)*b^3*x^3 - 5*(a^5 - 2*a^3 + a)*b^2*x^2 + 2*(a^6 - 3*a^4 + 3*a^
2 - 1)*b*x)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1))/((a^8 - 4*a^6 + 6*a^4 - 4*a^2 + 1)*x^4), -1/24*(3*(2*a^3 + 3*a
)*sqrt(a^2 - 1)*b^4*x^4*arctan(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(a*b*x + a^2 - 1)*sqrt(a^2 - 1)/((a^2 - 1)*b
^2*x^2 + a^4 + 2*(a^3 - a)*b*x - 2*a^2 + 1)) + 6*(a^8 - 4*a^6 + 6*a^4 - 4*a^2 + 1)*arcsin(b*x + a) - ((11*a^4
- 7*a^2 - 4)*b^3*x^3 - 5*(a^5 - 2*a^3 + a)*b^2*x^2 + 2*(a^6 - 3*a^4 + 3*a^2 - 1)*b*x)*sqrt(-b^2*x^2 - 2*a*b*x
- a^2 + 1))/((a^8 - 4*a^6 + 6*a^4 - 4*a^2 + 1)*x^4)]

________________________________________________________________________________________

giac [B]  time = 1.15, size = 1112, normalized size = 5.98 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(b*x+a)/x^5,x, algorithm="giac")

[Out]

-1/12*b*(3*(2*a^3*b^4 + 3*a*b^4)*arctan(((sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)*a/(b^2*x + a*b) - 1)/
sqrt(a^2 - 1))/((a^6*abs(b) - 3*a^4*abs(b) + 3*a^2*abs(b) - abs(b))*sqrt(a^2 - 1)) - (36*(sqrt(-b^2*x^2 - 2*a*
b*x - a^2 + 1)*abs(b) + b)^2*a^7*b^4/(b^2*x + a*b)^2 + 18*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^4*a^
7*b^4/(b^2*x + a*b)^4 + 18*a^7*b^4 - 81*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)*a^6*b^4/(b^2*x + a*b)
- 108*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^3*a^6*b^4/(b^2*x + a*b)^3 - 27*(sqrt(-b^2*x^2 - 2*a*b*x
- a^2 + 1)*abs(b) + b)^5*a^6*b^4/(b^2*x + a*b)^5 + 120*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^2*a^5*b
^4/(b^2*x + a*b)^2 + 81*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^4*a^5*b^4/(b^2*x + a*b)^4 - 5*a^5*b^4
+ 12*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)*a^4*b^4/(b^2*x + a*b) - 42*(sqrt(-b^2*x^2 - 2*a*b*x - a^2
+ 1)*abs(b) + b)^3*a^4*b^4/(b^2*x + a*b)^3 + 18*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^5*a^4*b^4/(b^
2*x + a*b)^5 - 18*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^2*a^3*b^4/(b^2*x + a*b)^2 - 36*(sqrt(-b^2*x^
2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^4*a^3*b^4/(b^2*x + a*b)^4 + 2*a^3*b^4 - 6*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 +
1)*abs(b) + b)*a^2*b^4/(b^2*x + a*b) + 8*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^3*a^2*b^4/(b^2*x + a*
b)^3 - 6*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^5*a^2*b^4/(b^2*x + a*b)^5 + 12*(sqrt(-b^2*x^2 - 2*a*b
*x - a^2 + 1)*abs(b) + b)^2*a*b^4/(b^2*x + a*b)^2 + 12*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^4*a*b^4
/(b^2*x + a*b)^4 - 8*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^3*b^4/(b^2*x + a*b)^3)/((a^9*abs(b) - 3*a
^7*abs(b) + 3*a^5*abs(b) - a^3*abs(b))*((sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^2*a/(b^2*x + a*b)^2 +
a - 2*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)/(b^2*x + a*b))^3)) - 1/4*arcsin(b*x + a)/x^4

________________________________________________________________________________________

maple [A]  time = 0.01, size = 309, normalized size = 1.66 \[ -\frac {\arcsin \left (b x +a \right )}{4 x^{4}}-\frac {b \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{12 \left (-a^{2}+1\right ) x^{3}}-\frac {5 b^{2} a \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{24 \left (-a^{2}+1\right )^{2} x^{2}}-\frac {5 b^{3} a^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{8 \left (-a^{2}+1\right )^{3} x}-\frac {5 b^{4} a^{3} \ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{b x}\right )}{8 \left (-a^{2}+1\right )^{\frac {7}{2}}}-\frac {3 b^{4} a \ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{b x}\right )}{8 \left (-a^{2}+1\right )^{\frac {5}{2}}}-\frac {b^{3} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{6 \left (-a^{2}+1\right )^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsin(b*x+a)/x^5,x)

[Out]

-1/4*arcsin(b*x+a)/x^4-1/12*b/(-a^2+1)/x^3*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-5/24*b^2*a/(-a^2+1)^2/x^2*(-b^2*x^2-
2*a*b*x-a^2+1)^(1/2)-5/8*b^3*a^2/(-a^2+1)^3/x*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-5/8*b^4*a^3/(-a^2+1)^(7/2)*ln((-2
*a^2+2-2*a*b*x+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/b/x)-3/8*b^4*a/(-a^2+1)^(5/2)*ln((-2*a^2+2-2*a
*b*x+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/b/x)-1/6*b^3/(-a^2+1)^2/x*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(b*x+a)/x^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a-1>0)', see `assume?` for mor
e details)Is a-1 positive, negative or zero?

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {asin}\left (a+b\,x\right )}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(a + b*x)/x^5,x)

[Out]

int(asin(a + b*x)/x^5, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asin}{\left (a + b x \right )}}{x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asin(b*x+a)/x**5,x)

[Out]

Integral(asin(a + b*x)/x**5, x)

________________________________________________________________________________________